20cm3 of 0.1m sulphuric acid was required to react with 50.0cm3 of a solution of sodium hydroxide
Very interesting but I don't see a question. If you wanted to know the molarity of the NaoH, you need to tell us what indicator was used; i.e.,did 1 or 2 H ions react with the NaOH?. Another point is did you mean 0.1m. That stands for MOLAL, not MOLAR.
2H IONS WAS USED AND ITS MOLAR
To determine the reaction between 20 cm³ of 0.1 M sulfuric acid and 50.0 cm³ of a solution of sodium hydroxide, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction.
Step 1: Write the balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH):
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Step 2: Convert the given volumes to moles using the molarity (M) of the sulfuric acid.
Molarity (M) = moles/volume (in liters)
Volume (in liters) = volume (in cm³) / 1000
Moles of sulfuric acid = Molarity × volume (in liters)
= 0.1 × (20/1000)
= 0.002 moles
Step 3: Use the stoichiometry of the balanced chemical equation to determine the moles of sodium hydroxide required for the reaction.
From the balanced equation, we see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.
Moles of sodium hydroxide = Moles of sulfuric acid × (2 moles of NaOH / 1 mole of H₂SO₄)
= 0.002 moles × (2/1)
= 0.004 moles
Step 4: Convert the moles of sodium hydroxide to volume (in cm³) using the molarity (M) of the sodium hydroxide solution.
Volume (in liters) = moles / Molarity
Volume (in cm³) = Volume (in liters) × 1000
Volume of sodium hydroxide = (0.004 / 0.1) × 1000
= 40 cm³
Therefore, 40 cm³ of the solution of sodium hydroxide is required to react with 20 cm³ of 0.1 M sulfuric acid.